Angular Mechanics I
by Jason "Mussolini" Crandall, Doug "Castro" Antoni,
Martin "The Little Helper" Held
and Keith "Yoda" Banks, January 1998
Table of Contents
Here are the quantities you will soon be able to calculate:
Ax Rotational Displacement
t Time
f Frequency
Av_{o} Initial Angular Velocity
Av Final Angular Velocity
Av_{a} Average Angular Velocity
Aa Angular Acceleration
a_{c} Centripetal Acceleration
m mass
Tq Torque
I Moment of Inertia
r Radius
l length
None of these quantities are defined on other pages, so we will explain them all.
Please note that many of the linear quantities you are used to working with are
not listed here, but you will be expected to know them.
New Quantities
Defining Angular Displacement
You already know what linear displacement is. Linear Displacement is the distance the object moves
But what if the object is spinning, such as a wheel? This is angular displacement. Angular displacement is the change in position of a point on a object. Linear displacement is measured in Meters (m). Angular displacement is measured in radians. DO NOT confuse Angular and Linear Displacement.
A simple example here will illustrate the difference between Linear and Angular
displacement. Suppose that a Frisbee is thrown at 40 m/s(approximately 90 mi/h)
Spinning once every two minutes due to a weak wrist motion.
To find the linear displacement after one minute, all we have to do is look at the Frisbee's speed. It goes 2400 meters in that one minute. Now, what is the Frisbee's angular displacement? Well, we know that it turns once every two minutes. This is .5 Rpm (rev per minute). Problem. We are looking for radians per minute, how to find this?
Hopefully, you know what radians are, if not, then you need more help then you can get here. The angular displacement of the Frisbee during this time is .5 rpms * 2(pi).
Our new formula is:
(Change in displacement)=(Angular velocity (RPMs)) * (2(Pi))
Defining Angular Velocity
We briefly included angular velocity in the previous section. Angular velocity is
basically analogous to linear velocity. It is the change in position of a rotating point
over time. However, instead of distance traveled, we are now describing radians
rotated. The symbol for angular velocity is Av or omega. To use the formulas we
have, merely replace the v with Av and x with Ax. Angular velocity is measured in
rad/s. We will discuss angular velocity further in the example problems.
Defining Angular Acceleration
Like many of the other terms that we have seen so far, angular acceleration
is quite like linear acceleration. It is simply the rate of change in
rotation rather than the rate of change in velocity. For linear velocity,
we used meters per second, and we then used meters per second squared( m/s^{2})
for acceleration. Of course, we use radians per second for Angular Motion,
and you guessed it radians per second squared( rad/s^{2}) for
Angular Acceleration. Many of the same equations from linear kinematics apply
here as well, and you need only replace a(acceleration) with Aa for Angular Acceleration.
Defining Torque
Torque is the angular equivalent of force, it is what causes rotational motion. The difference is that Force imparts linear acceleration; whereas, torque gives an object angular acceleration. F=M*A and T=F*D, where D is the Force's distance from the center of rotation force multiplied by distance yields the units Newton Meters. Torque is the reason why the further out on a seesaw you are the more influence you have because as you move further back your distance from the axis of rotation increases. The problem is that torque is based on the force acting perpendicularly. To account for this one must multiply the distance by the sine of the angle from the horizontal. So if 100 Newtons of force were to act on a .2 Kg pinwheel at a 45 degree angle 10 cm from the center. The Torque would be 100N*.1M*Sin(45)= 7.07106 NM. To convert this answer to rotations per minute we would have to use a moment of inertia, but to find it would require calculus, so I will declare it to be I=.5*M*R^2. Then using the formula T=I[A] we find
that [A]=7071.1 rad/S. Since there are 2(Pie) in each rotation dividing our answer by it would yield 1125.39 rpm.
Example: A force of 10N is exerted on the head of a rubber dummy that is 15 kg in weight, 2 meters tall, and is attached to the floor. What is the Torque exerted on the object and what is [A]?
Torque= Force*Distance so 10*2= 20 N of Torque
Torque= 20 N so because Newtons are Mass*Acceleration, Torque=Distance*Mass*Angular Acceleration
20/15 = 1.333 so [A]=1.333 rad/s*s
Defining Moment of Inertia
This one sounds pretty rough, but its not all that bad. If we continue our comparison
between linear and angular formulas, we've already defined the angular equivalent of
distance, velocity, acceleration and force. But what about mass?
Mass in term of rotational motion is called the moment of inertia. This quantity comes
from the linear equation F=ma. By substituting angular acceleration into this equation
using the relationship between tangential acceleration and angular acceleration we
arrive at:
F = m Aa r(where r is the radius)
We then multiply both sides by r and find:
Fr (Tq) = m r^{2} Aa
So now, instead of force equals mass times acceleration, we have torque equals mr^{2}
times angular acceleration. This excrescence is referred to as the moment of inertia or
rotational inertia. The moment of inertia, or I, is written in units of kgm^{2}. However,
the equation for moment of inertia differs depending on the shape of the rotating
object. For example, the moment of inertia of a ring and cylinder with equal mass
and radius is very different. This is because the distribution of mass in the cylinder
is uniform across the entire surface, while there obviously cannot be any mass in the
center of the ring. Although it is very difficult to determine the distribution of mass
for many objects, here is a list of common objects and their moments of inertia.
Moment of Inertia Table
(a) Thin ring = m r^{2}
(b) Uniform cylinder = 1/2 m r^{2} (rotated through center)
(c) Uniform sphere = 2/5 m r^{2} (rotated through center)
(d) Long uniform rod (rotated through center) = 1/12 m l^{2} (l is the length of the rod)
(e) Long uniform rod (rotated about an end) = 1/3 m l^{2}
Go back to: Example Problem 4 Example Problem 5
Sample Problems
Tangential Relationships
This has absolutely nothing to do with hormones. Really. I'm serious. What I'm about to tell you might be difficult to grasp, but you will understand it, or I will kill you. (No, I won't. Just kidding.... maybe. ) When we say tangential relationships, we are talking about the differences between angular quantities and linear quantities, which we hope you have noticed by now. The first rule to remember is this: Send your life savings to Martin Held, 150... Oh wait. Wrong rule. The REAL first rule is that you CANNOT confuse any angular quantity with any linear quantity. Now, because of this rule, we will have to convert a lot of numbers when going from angular to linear and visaversa. But first, we must learn something new (or maybe not so new): The word "tangential". Tangential means, for us, "On the outer edge". When someone talks about the "tangential velocity" they are talking about the linear velocity of a point on the outer edge of a circle with radius r, as it goes around, and around, and around, and around, and around, and around, and...
*puke* *puke* *retch*
OK. You get the idea. Just to make your life, there are 6 major things that you will have to be able to do: convert linear displacement to angular displacement, convert linear velocity to angular velocity, and convert linear acceleration to angular acceleration. What are the other three? Exactly the opposite. Convert angular displacement to linear displacement, yadda yadda yadda. On to the methods!
First, to convert angular and linear displacements, you are going to have to know the radius of the circle so that you can find the circumference, and then you can convert by dividing the number of radians it turned by 2(pi) and multiplying by the circumference. This will tell us how many revolutions it turned, and thus how many times it turned through the circumference. Example:
A wheel of radius 1 meter just turn through 2 radians. What was the tangential displacement of the wheel?
To find this, we need to find the circumference, which is 2(pi) ( I DO hope you remember how to find the circumference by using 2(pi)r !!! ). We then find the number of revolutions by diving 2 by 2(pi) which gives us 1/(pi). When we multiply the result by the 2(pi) (a.k.a. the circumference) to get 2(pi)/(pi), which simplifies to 2 meters.
Now that you know this, the rest of the conversions are actually quite easy. They all involve dividing radians, radians per second, and radians per second squared by 2(pi), and then multiplying by the circumference. To convert from linear to angular, just do the reverse operation. Example:
The outer edge of a .5 m radius wheel undergoes a 20 meter/sec/sec acceleration. What is the angular acceleration of the wheel in radians per second squared?
Easy. Circumference = 2 * (pi) * .5 = (pi).
(pi)/20 = .157 rev/sec/sec
((pi)/20) * (2(pi)) = (pi^{2})/10 = .98696 rad/sec/sec.
See how easy it is? Now get out there and try some of those impossible problems, courtesy of Jason Crandall.
Formulas
Here is a list of the all of the formulas we will be using:

Av = 2(pi)f

Av = Av_{o} + Aa t

Av_{a} = (Av_{o} + Av)/_{2}

Ax = Av_{a} t

Ax = Av_{o} t + ^{1}/_{2} Aa t^{2}

a_{c} = Av^{2} r

Tq = I Aa

I = m r^{2}

Aa=( Av  Av_{o} ) / t
Go back to: Table of Contents
General Problem Solving Strategy:
 Read the problem.
 Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.  Find any formula that will allow you to calculate
anything that you don't know, and apply it.  Add what you just found in the last step to your list of knowns.
 Check to see if you have found the answer. If not, repeat the
previous two steps until you are done.
Go back to: Table of Contents
Example problem 1
Little Johnny is at the playground. Johnny really likes the merrygoround. Johnny asks his brother to push the merrygoround. a) If Johnny's brother pushes the merrygoround so that it goes from rest to a speed of 50 rad/sec in eight seconds, what is the angular acceleration? b) If Johnny flies off the 6 meter diameter merrygoround while it is going 50 rad/sec, with what speed does Johnny leave the merrygoround? c) Should Johnny file criminal charges?
First, we need to refer to our list of formulas and find the appropriate formula for solving angular acceleration. Here is what we know:
Av_{o} = 0 rad/sec
Av = 50 rad/sec
t = 8 sec
Aa = ?
Now we use the formula, Aa=( Av  Av_{o} ) / t, and solve for angular acceleration. Here is the formula with the numbers plugged in,
Aa = (50  0)/8.
So the answer to a) is 6.25 rad/s/s.
To solve part B, we must use tangential relationships. The relationship between tangential velocity and angular velocity is quite simple. Just multiply the angular velocity by the radius of the spinning object (the merrygoround, not Johnny). The merrygoround has a six meter diameter so its radius is 3 meters. Fifty rad/sec times 3 meters is 150 m/s or 335 miles per hour.
As for part C, this is pretty much a judgment call. I'm sure Johnny's brother didn't mean to hurt him...did he?
Go back to: Table of Contents
Example problem 2
Little Johnny needs major brain surgery. To open his skull,
a professional Craftsman drill must be used. The drill will drill
down a certain amount with every rotation, so it is important to
know how long it will take for the drill to go in a certain
distance. The drill going 260 rad/sec goes through 25 radians in
how much time? If the drill goes 3.7 inches, what happens
to Johnny?
Here is what you start with:
t = ?
Av = 260 rad/sec
Ax = 25 radians
So, if you look at the formulas, the third one will work because we
are given the angular velocity and angular displacement. The only
unknown quantity in this formula is the time, which is what the
question asks for. So:
Ax = Av t
putting in numbers:
25 radians = (260 rad/sec)* t
so now solve for t:
t = .096153846154 s
Now we know everything:
t = .096153846154 s
Av = 260 rad/sec
Ax = 25 radians
Since the question asks for time, we correct for significant digits (3
in this case) and arrive at the solution:
Time = .096 seconds.
After extensive research on the thickness of children's skulls, we've
determined that the procedure needs to be rethought (and the sick
freak who wrote this problem needs to be committed).
Go back to: Table of Contents
Example problem 3
The drill must be powered up before it will start drilling into
Johnny's head. This time must be taken into account to make sure
Johnny doesn't get the wrong hole in his head. The drill can speed
up at 45 rad/sec^{2}. What time will it take to speed up from 130
rad/sec to 260 rad/sec?
Here's what we know:
V_{o} = 130 rad/sec
V = 260 rad/sec
Aa = 45 rad/s/s
If you look at the formulas:

Av = Av_{o} + at

Av_{a} = (Av_{o} + Av)/_{2}

Ax = AV_{a}t
The only one that we can use is the first. Plugging numbers into the
formula:
260 rad/sec = 130 rad/sec + (45 rad/s/s*t)
So t = 2.88 s. So now we know:
t = 2.88 s
V_{o} = 45 rad/sec
V = 120 rad/sec
a = 23 rad/s/s
The problem asks you to find the time which is 2.88 s so we're through
Go back to: Table of Contents
Example problem 4
Little Johnny has recovered from his brain surgery. But he's still
in the hospital. He wants to learn something new that doesn't require too much energy.
His mother suggest hulahooping. Johnny realizes that his mother is still a hippie but
decides to give it a shot. In order for Johnny to determine how little work he can do
while hulahooping, he must figure out the moment of inertia of his hoop. What is the
moment of inertia of a .5 Kg, 51 cm radius hula hoop about Johnny's waist?
The first thing we need to do is check the Moment of Inertia Table. We see that the
moment of inertia for a thin ring (such as a hula hoop), is I = m r^{2}. We know:
m = .5 kg
r = 51 cm
I = ?
So we just have to plug in m and r:
I = .5 * .51^{2}
Therefore, I equals .13 kgm^{2}. Of course, this does not guarantee that Johnny will ever be able to keep the hoop up.
Go back to: Table of Contents
Example problem 5
Johnny is very tired from hulahooping all day long. However, he can not get into bed because it is stuck. a) What is know the angular acceleration of his 1.9 meter long, 225 kg bed if Johnny exerts a torque of 70 Nm? b) What is the tangential force which Johnny exerts?
We've got to find the moment of inertia before we can do anything. We can assume from the shape of most beds that Johnny's resembles a uniform rod. After checking the Moment of Inertia Table, we see that a uniform rod rotated about its end has a moment of inertia of 1/3 m l^{2}. We plug in the mass and the length of Johnny's bed
I = 1/3 (225) 1.9^{2}
and find that the moment of inertia to be 270.75 kgm^{2}. Now we use the formula Tq = I * Aa, and find that Johnny's bed accelerates at .26 rad/s/s when he exerts a torque of 70 Nm.
The tangential force which Johnny exerts can be found by the formula Tq = F r (or, in this case, Tq = F l). Once again, the length of the bed is 1.9 meters and the torque is 70 Nm. Divide 70 by 1.9 and you get 36.8 N of tangential force.
Go back to: Table of Contents
Sample Problems
1.
A wheel turns 34 radians in 3.2 seconds. What's its angular velocity?
(10.6 rad/sec)
2.
A drill going 340 rad/sec goes through 12 radians in how much time?
(.035 seconds)
3.
A motor goes from 12 rad/sec to 68 rad/sec in .12 seconds. What is the
angular acceleration?
(467 rad/sec/sec)
4.
A fan can speed up at 23 rad/sec^{2}. What time will it take to speed up from
45 rad/sec to 120 rad/sec?
(3.26 sec)
5.
A pulley speeds up from 34 rad/sec to 80 rad/sec in 12 seconds.
a) what is the angular acceleration?
b) what is the average angular velocity?
c) what is the angle the pulley goes through?
(3.8 rad/s/s, 57 rad/sec, 684 rad)
Go back to: Table of Contents
6.
A merrygoround slows to rest from a speed of 34 rad/sec over a time
of 190 seconds.
a) What is the angular acceleration?
b) What is the average angular velocity?
c) What angle does it go through in this time?
(a. .18 r/s/s, b. 17 r/s, c. 3230 r )
7.
A 12 cm radius grinding wheel is rotating at 160 rad/sec.
What is the tangential velocity at the edge of the wheel?
(19.2 m/s)
8.
A car accelerates at 8.2 m/s/s. What is the angular acceleration
of the wheels if they have a radius of 35 cm?
(23.4 r/s/s)
9.
What angle does a 35 cm radius car wheel go through if the car travels
a distance of 12 m?
(34.3 r)
10.
A rollerblade wheel is 3.2 cm in radius.
a) What is the angular velocity of the wheel if you are going 14 m/s?
b) What is that in revolutions per second?
c) What is that in revolutions per minute?
(437.5 r/s, 69.6 rev/sec, 4180 rpm)
Go back to: Table of Contents
11.
A .12 m radius disk sander has a motor that spins the disc at
1200 rpm.
a) What is the angular velocity in rad/sec?
b) What is the tangential velocity of the disk at the edge of it?
(a. 126 r/s, 15.1 m/s)
12.
A carnival ride slows from 1.2 rev/sec to rest in 5.2 seconds.
The riders sit at a radius of 2.1 meters from the center of rotation.
a) What is the initial angular velocity in rad/sec?
b) What was the angular acceleration?
c) Through what angle did the ride turn during this time?
d) What was the tangential acceleration of the riders?
e) What distance did the riders travel in this time?
(7.54 rad/s, 1.45 rad/s/s, 19.6 rad, 3.04 m/s/s, 41.2 M)
13.
A drill speeds up from 1200 rpm to 3400 rpm in .12 seconds.
a) What is the initial angular velocity in rad/sec?
b) what is the final angular velocity in rad/sec?
c) What is the average angular velocity in rad/sec?
d) What angle does the drill turn through in that time?
e) If there is a drill bit with a diameter of .0062 m on it,
what is the final tangential speed of the bit?
(a. 126 r/s, b. 356 r/s, c. 241 r/s, d. 28.9 r, e. 2.21 m/sec)
14.
What is the moment of inertia of a 5 kg 34 cm radius hoop about its center?
(.578 kgm^{2})
15.
What is the moment of inertia of a 6 Kg 2.1 m long walking stick
about its center?
(2.205 kgm^{2})
16.
What is the moment of inertia of a 8 kg 10 cm radius sphere
about its center?
(.032 kgm^{2})
17.
What is the moment of inertia of a 100 kg 1.8 m radius solid cylinder
about its center?
(162 kgm^{2})
18.
What is the torque when you exert a force of 52 N on a 56 cm breaker bar?
(29 Nm)
19.
What force should you exert on a 14 cm long wrench to get a torque
of 42 Nm?
(300 N)
20.
If you exert a torque of 68 Nm on a flywheel with an I of 12 Kgm2,
what is its angular acceleration?
(5.7 rad/s/s)
21.
A drill exerts a torque of 80 Nm on a 1.2 Kg .12 m radius grinding
disk that is a solid cylinder. What is the angular acceleration of
the disk?
(9259 r/s/s)
Go back to: Table of Contents
Solutions to Sample Problems
1.
A wheel turns 34 radians in 3.2 seconds.
What's its angular velocity?(10.6 rad/sec)
Here is what you start with:
Av = ?
Ax = 34 radians
t = 3.2 sec
Let's look back at the formulas. The third one will work for this simple problem. You get:
Ax = Av t
putting in numbers:
34 radians = Av* 3.2
so solve for Av:
Av = 10.6 rad/sec
Go to: Problem Formulas Table of Contents
2.
A drill going 340 rad/sec goes through 12 radians in how much time?
(.035 seconds)
Here is what you start with:
t = ?
Av = 340 rad/sec
Ax = 12 radians
So, if you look at the formulas, the third one will work because we
are given the angular velocity and angular displacement. The only
unknown quantity in this formula is the time, which is what the
question asks for. So:
Ax = Av t
putting in numbers:
12 radians = (340 rad/sec)* t
so now solve for t:
t = .035 s
Go to: Problem Formulas Table of Contents
3.
A motor goes from 12 rad/sec to 68 rad/sec in .12 seconds. What is the angular acceleration?(467 rad/sec/sec)
Here is what we know:
t = .12
Av_{0} = 12 rad/sec
Av = 68 rad/sec
Aa = ?
So know lets take a look at our formulas, shall we?Formulas
As you can see, we want the formula:
Av = Av_{0} + Aa t
Plugging in numbers:
68 rad/sec = 12 rad/sec + Aa .12
Solve for Aa and you get:
467 rad/sec/sec
Go to: Problem Formulas Table of Contents
4.
A fan can speed up at 23 rad/sec^{2}. What time will it take to speed up from
45 rad/sec to 120 rad/sec?
(3.26 seconds)
Here's what we know:
V_{o} = 45 rad/sec
V = 120 rad/sec
a = 23 rad/s/s
If you look at the formulas:

Av = Av_{o} + at

Av_{a} = (Av_{o} + Av)/_{2}

Ax = AV_{a}t
The only one that we can use is the first. Plugging numbers into the
formula:
120 rad/sec = 45 rad/sec + (23 rad/s/s*t)
So t = 3.26 s. So now we know:
t = 3.26 s
V_{o} = 45 rad/sec
V = 120 rad/sec
a = 23 rad/s/s
The problem asks you to find the time which is 3.26 s so we're through.
Go to: Problem Formulas Table of Contents
5.
A pulley speeds up from 34 rad/sec to 80 rad/sec in 12 seconds. a) what is the angular acceleration? b) what is the average angular velocity? c) what is the angle the pulley goes through?
First, we will use the equation Aa=( Av  Av_{o} )/t to find the angular equation. You've seen one like this before, right? We started at 34 rad/sec, and ended at 80 rad/sec, so:
Av_{o} = 34 rad/s, Av = 80 rad/s, and t = 12. 46/12 = 3.8 rad/sec^{2}.
To find the average angular velocity, we use the average formula, (Av_{o} + Av)/2. (34+80)/2 = 57 rad/s.
Finally, to find the angle the pulley goes through, we use the average velocity and multiply by the time, to find the number of radians it turned in 12 seconds. 57 * 12 = 684 radians.
Go To:
Problem Formulas Table of Contents
6.
A merrygoround slows to rest from a speed of 34 rad/sec over a time
of 190 seconds.
a) What is the angular acceleration?
b) What is the average angular velocity?
c) What angle does it go through in this time?
(a. .18 r/s/s, b. 17 r/s, c. 3230 r)
All right, here we go. To find the angular acceleration, use the
formula Aa = Av  Av_{o}/t. With numbers plugged in, it looks like
0 rad/sec34 rad/sec / 190 sec. Aa equals .18 r/s/s. For part b, we
use a similar formula, Av_{a} = Av + Av_{o}/2. We know Av (0 rad/sec) and
Av_{o} (34 rad/sec), so Av_{a} is 17 r/s. Finally, to find the angle, we use
an angular displacement formula. We know:
t = 190
Av = 0 rad/sec
Av_{o} = 34 rad/sec
Av_{a} = 17 rad/sec
Aa = .18 rad/s/s
Ax = ?
So we can use formula #4 or #5. Four is considerably simpler so I will
be using for this example. Ax = Av_{a} t. Therefore, Ax is 3230 radians.
Go to: Problem Formulas Table of Contents
7.
A 12 cm radius grinding wheel is rotating at 160 rad/sec. What is the
tangential velocity at the edge of the wheel? 19.2 m/s
If you will recall the section on tangential relationships (written
so poetically by Martin Held), you know that this is a very easy
problem. Merely multiply 160 rad/sec by .12 m because the tangential velocity is the angular velocity times the radius. Our
answer is 19.2 m/sec.
Go to: Problem Formulas Table of Contents
8.
A car accelerates at 8.2 m/s/s. What is the angular acceleration
of the wheels if they have a radius of 35 cm?(23.4 r/s/s)
This problem is a simple tangential relationship problem. Here is what we know:
a = 8.2 m/s/s
Aa = ?
r = .35 m
This problem is so simple, that if you look back at the tangential velocity area (again, written by Martin Held) you merely divide the linear accelaration (8.2 m/s/s) by the radius of the tires (.35 m) and you get:
23.4 r/s/s
Go to: Problem Formulas Table of Contents
9.
What angle does a 35 cm radius car wheel go through if the car travels
a distance of 12 m?(34.3 r)
Here is what you start with:
r=.35 m
x= 12 m
C = ?
Theta= ?
Use C= 2*Pi*r to find the circumference: 2.199114 m. Divide the
distance traveled by that answer which equals the revolutions made,
5.46 revolutions. Then multiply that by 2 * Pi and your answer is at
hand, 34.3 radians.
Go to: Problem Formulas Table of Contents
10.
A rollerblade wheel is 3.2 cm in radius. a) What is the angular velocity of the wheel if you are going 14 m/s? b) What is that in revolutions per second? c) What is that in revolutions per minute?
First, we are going to need to find out what the circumference is. 2(pi)r will give us the answer, so we get 20.10 cm, which is .2010 meters. We know that the outer edge of the wheel has to go 14 meters per second, and so we divide 14 by .2010 meters = 69.65 revolutions per second. Whoops! that's question b! Either way, we can still go on from here. We simply need to convert revolutions per second to radians per second by multiplying by 2(pi), and we get 437.6 radians per second. Now we only have to find revolutions per minute. We know that it is running at 69.65 rev/sec, so we multiply by 60 to find revolutions per minute, and we get 437.6 revolutions per minute. Yippee.
Go to: Problem Formulas Table of Contents
11.
A .12 m radius disk sander has a motor that spins the disc at 1200 rpm.
a) What is the angular velocity in rad/sec?
b) What is the tangential velocity of the disk at the edge of it?
Here is what you start with:
r=.12m
rpm= 1200
Using 1200 rpm = 20 rps then multiplying by 2 * Pi you receive the
answer, 126 r/s.
Part b uses the formula: C= 2*Pi*R the circumference is: 1.11937m.
So after converting 1200 rpm to 20 revolutions per second just
multiply 20*1.11937 to get the answer: 22.3874 m
Go to: Problem Formulas Table of Contents
12.
A carnival ride slows from 1.2 rev/sec to rest in 5.2 seconds. The riders sit at a radius of 2.1 meters from the center of rotation. a) What is the initial angular velocity in rad/sec? b) What was the angular acceleration? c) Through what angle did the ride turn during this time? d) What was the tangential acceleration of the riders? e) What distance did the riders travel in this time?
Here's what we know:
f_{o} = 1.2 rev/sec
AV = 0
Va = ?
It looks like we have many unknowns here, but that is not a problem. We'll just work from the beginning and find out what we need to know as we go.
To start, we need to find out AX. To do this, we use f, which is given. The formula is: Av = 2(pi)f
Plugging in Numbers, we get:
Av = 2(pi)1.2 rev/sec
So:
Av = 7.53982236862
Now we need to find the angular acceleration as directed in part B.
Do to this, we use Av = Av_{o} + Aa t
We plug in the numbers given and the answer from part A as AV and get:
0=7.5398 + Aa 5.2
Solving for Aa you get: 1.45 rad/s/s (It's going backwards)
Now for part C. To do this, we have to use two formulas, Av_{a} = (Av_{o} + Av)/_{2}, and Ax = Av_{a} t. This finds use the average Angular Velocity and then the Angular displacement. Plugging in numbers:
Av_{a} = (0 + 7.5398)/_{2} SO:
Av_{a} = 3.7699 Then, we take this answer and put it in the second formula:
Ax = 3.7699 5.2 So:
Ax = 19.6
Now for part d, we need to find the tangential acceleration of the riders on the ride. First, we assume that they are on the outside edge of the ride. This makes it simpler. As you learned in the Tangential Relationships area, we need the circumference of the ride. To find this, we simply use the formula you learned in Geometry:
Circumference = 2(pi)R
Plugging in the radius of 2.1 meters, you get an answer of:
Circumference = 13.3 meters
Now, we need to turn this into the initial Tangential Velocity. To do this, we need to multiply it by the revolutions per second, in this case, 1.2. So:
Tang. Vel. = 15.8 m/s
Now, all we need to do is divide by the time, 5.2 s, throw in a negative sign because they are slowing down, and we have our answer, so:
tangential acceleration = 3.04 m/s/s
Finally, we on part e, the last part in this long problem. In this part, we need to find the distance the riders travel in those magical 5.2 seconds everyone is always talking about. This is simply an Angular Displacement problem with a conversion at the end.. First, we need to find the average angular velocity. So, we simply add the initial and final velocities:
Av_{a} = (7.54 rad/s + 0)/_{2}
So: Av = 3.77 rad/s
Now, we take that answer and plug it into the formula:
x = ((Ax t)_{/2(pi)})C
Where C is the circumference
We plug in the numbers and get the answer:
41.2 Meters
Now we are done with this problem.
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13.
A drill speeds up from 1200 rpm to 3400 rpm in .12 seconds.
a) What is the initial angular velocity in rad/sec?
b) what is the final angular velocity in rad/sec?
c) What is the average angular velocity in rad/sec?
d) What angle does the drill turn through in that time?
e) If there is a drill bit with a diameter of .0062 m on it, what
is the final tangential speed of the bit?
To find the speed in rad/sec, we need to convert the speed in
revolutions per minute to revolutions per second, then to radians
per second. We know that because the drill will turn 1200
revolutions in one minute, it must complete 1/60th of this every
second. Knowing this, we divide 1200 revolutions per minute by
60 to get 20 revolutions a second. If we want to double check this,
we know that it will do 20 radians every second for 60 seconds,
and thus 1200 rev/min. Wasn't that FUN???. Now that we know
how fast it goes in revolutions per second, and we can tell that
there are 2(pi) radians for every revolution, so we simply multiply
revolutions per second by 2(pi) to radians per second.
20 * 2 * (pi) = 126 rad/sec.
Next to find the final angular velocity, we can do the same thing
to find the final velocity as we did the initial velocity.
(3400 / 60 ) = 56.6
56.6 * 2 * (pi) = 356 rad/sec.
Because we now know the initial and final velocity of the drill, it is
easy to find the average. All we have to do is add the two together,
and divide by two.
(126 + 356 ) / 2 = 241 rad/sec.
Now that we know the average angular velocity of the drill and how
long it turned at that average rate, we can simply multiply the two
together to find the total angular displacement.
241 * .12 = 28.92 radians.
Now comes the tricky part. We have to find the average velocity of
the outer edge of the drill bit, but to do this, we must only do one
thing: convert radians per second to meters per second. First, we
have to reverse the process that we used in parts a and b to find the
speed of the drill in revolutions per second (we do not need to know
the speed in revolutions per minute, however). To do this, we divide
our average angular speed of 241 rad/sec by 2(pi), and we get 38.35
revolutions per second.
241 / ( 2 * pi ) = 38.35
The next step is to find out the circumference of the bit using 2(pi)r.
2 * (pi) * .0062 = .038955 meters
We know that it will have to turn 38.35 times every second, and
every time it turns once, a point on the outer edge travels through
the circumference, and thus, the circumference is traversed by such
a point 38.35 times every second. If we multiply the circumference
by the speed of the drill bit in revolutions per second, we will get
the number of meters traveled every second, i.e. meters per second,
the velocity we are looking for
.038955 * 38.35 = 1.49 meters per second.
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14.
What is the moment of inertia of a 5 kg 34 cm radius hoop about its center?(.578 kgm^{2})
As with all Moment of Inertia problems, we will use the formula:
I = m r^{2}
Looking back at the top, we see that, since this is a hoop, it shares the same properties as a thin ring, and therefore the formula is unchanged. So, we plug in the numbers and get:
.578 kgm
Moment of Inertia Table
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15.
What is the moment of inertia of a 6 Kg 2.1 m long walking stick
about its center? 2.205 Kgm^{2}
This one may seem hard at first, but its really quite simple. In
this problem, we need only find one thing, I. To do this, we
look back at the moment of inertia section at the top and see that
a long uniform rod (rotated through center) = 1/12 m l^{2} (L is the
length of the rod).
All we need to do is plug in the numbers for the mass and length.
Once we do this, it is a simple number punch in the calculator to
get 2.205 kgm^{2}.
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16.
What is the moment of inertia of a 8 kg 10 cm radius sphere
about its center?(.032kgm^{2})
Again, we use the Moment of Inertia formula. Since you have seen it enough, we will not show it again, we think you can remember. However, this problem is slightly different. If you look on the Moment of Inertia Table you will see that for a sphere you must then multiply the formula by 2/5 to get the right answer. (No, we dont know why, so don't ask) Punch a few numbers and you get:
.032kgm^{2}
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17.
What is the moment of inertia of a 100 kg 1.8 m radius solid cylinder
about its center?(162 kgm^{2})
Avast! What be ye? A Moment of Inertia problem ye be. By now, you have these things licked, and have probably skipped to the next section. If not, here goes...
Again, we use our friend:
I = M r^{2
And, looking at the Moment of Inertia Table we see that we must use 1/2 m r2 for a cylinder. Pulg in the numbers and you get:
162 kgm2
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18.What is the torque when you exert a force of 52 N on a 56 cm breaker bar?(29 Nm)
This is a easy problem. We must use the equation:
T = F D
Plugging in the numbers our answer is a hand:
29 Nm
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19.
What force should you exert on a 14 cm long wrench to get a torque
of 42 Nm? (300 N)
You shouldn't have too much trouble with this one. Just remember
that Torque = Force * Radius. In this problem, we are given the
torque and the radius, but not the force. So plug in your numbers
and solve for force, 42 = Force * .14. Force equals 300 N.
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20.If you exert a torque of 68 Nm on a flywheel with an I of 12 Kgm2, what is its angular acceleration? (5.7 rad/s/s)
To find this one, we look back up at the equations to find the one we need:
Tq = I Aa
Simply plug the numbers into their correct locations:
68 Nm = 12 Kgm2 Aa
And you get:
5.7 rad/s/s
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21.A drill exerts a torque of 80 Nm on a 1.2 Kg .12 m radius grinding
disk that is a solid cylinder. What is the angular acceleration
of the disk?(2.75 s)Here is what you start with:
t= 2.75s
Tq=80 NM
M= 1.2 Kg
r=.12 m
Using Tq= r*M*Aa and solving for Aa the answer is: 9259 r/s/s
You have done well, my son. You power in the force has grown strong. You must now go out and face your enemy, Darth Final!!
Please leave now, the tutorial has ended. I mean it. Leave.
Go, Now!! Ah, what do I care?
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